I have been trying to view this problem for a while and I have come unstuck, because of the at least at least part.
Calculate the probability that at least one of two students A and B, who independently attend each lecture of physics with a probability of 0.9 will fail to be present during at least 1 lecture of the 10 lecture series.
So I have tried to think about this several ways. Here is a few.
I first started to look at the probability of at least one of the students turning up to a lecture.
So this can be done two way.
1) I can look at the probability of both students turning up
$P(A\: and \: B)=\frac{81}{100}$ and then looking at the individual $P(A\: not \: B)=\frac{9}{100}$ and $P(B \: not \: A)=\frac{9}{100}$ so I can sum all these to get the probability of atleast 1 to be $P(At\:least \: 1)=\frac{99}{100}$
or
2) I can use the venn approach where $P(A \cap B)=P(A)+P(B)-P(A \cup B)=\frac{99}{100}$ if I sub in the given probabilities.
So then I thought if I could look at say two lectures and I get both methods to agree then I must be on to the right track. The issue is I can get either method to agree.
So I used a spread sheet as displayed below to calculate the probabilities for two lectures where A and B turn up individually or as a pair.
The last number is the sum of all the probabilities so then I make the probability of A and B being at least 2 lectures is $0.9639$.
I then used the following to confirm this $P(Atleast \: 1)=(0.9)^2+(0.9)^2-(0.81)^2=0.9639$
Now the two results do confirm however I am not quite convinced that it is correct at is you did it for 10 lectures the I have a probability that of $(Atleast \: 1)=0.58$.
I then did a little research and could not find anything that would confirm my results or how I have carried out the calculation. So I then came across the Bernoulli method where you have a given set of experiments and the probability is the same for each trails i.e like flipping a coin, so I decided to try and apply this method to the problem.
so according to the Bernoulli method I make the probability $(Atleast \: 1)=1-(1-\frac{1}{100})^10=0.0951$ which dose not line up with the preceding.
Could someone may shed light on where I am going wrong, and explain how I should approach.
Best Answer
You are starting out okay, but you're making the problem more difficult than it really is. When you have trouble calculating a probability it's often worthwhile whether it would be easier to calculate the complementary probability: the probability that the event does not occur. Then you just subtract from $1$ to get the probability you're really interested in.
The negation of "At least one of the students is not present at at least one of the lectures," is "Both students are present at all the lectures." Since you've already calculated that the probability that both are present at a given lecture is $.81,$ I'm sure you'll see how to proceed from here.